A camera is being used with the correct exposure at f/4 and a shutter speedof 1/

Question

A camera is being used with the correct exposure at f/4 and a shutter speedof 1/32s. In order to "stop" a fast moving subject, the shutter speed is changed to 1/256s. Find the new f stop that should be used to maintain satisfactory exposure, assuming no change in lighting conditions. The Answer is f/1.4.

I understand the first part. (1/256)/(1/32) so we know the exposure is reduced by a factor of 8 (8 times larger aperature).

I also know that this involves the f number = f/D formula.   D is the diameter of the lens. The rest of the teachers explaination don't make sense (he skips multiple steps a lot). So I don't understand how to take that I know the change will be 8 fold and produce the f/1.4 answer. Thanks for the help!

What is listed is

D2 = sqrt8 (D1)

f=f/D so f2 = f/D2= f/sqrt8(D1) = f1/sqrt8.

so f2 = 4sqrt8= 1.41 so f/1.4 is the setting. I want to be able to work this myself but can't follow this to do it. Thanks!

Explanation / Answer

The exposure time drops from 1/32 to 1/256. this means that there will be (1/32)/(1/256) = 8 times less light hitting the film.

So the open area of the lens must be 8 times larger than before. And because the area is

pr^2, the radius, or the diameter must be ?8 times larger than before.

And that is achieved when the new f number is 4/?8 = 1.414 (i.e., 1 should be ok)

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